Limit of a sum of natural logarithms

Question

As the title says I have to calculate a limit:

$$\lim_{x\to0}\left(\ln(1+x)+\ln(1+2x)+...+\ln(1+px)\right)^x$$

I've transformed the sum into one logarithm $\ln\left((1+x)(1+2x)...\right)$etc but I don't see how it helps me further. Some hints would be great.

Answer 1

Using $$\ln(1+x)\sim_0 x$$ we have

$$\left(\ln(1+x)+\ln(1+2x)+\cdots+\ln(1+px)\right)^x\sim_0\left(\frac{p(p+1)}{2}x\right)^x\\=\exp\left(x\ln\left(\frac{p(p+1)}{2}x\right)\right)\xrightarrow{x\to0}1$$

Answer 2

Hint: $\ln(1+t)\sim t$ as $t\to0$.

Answer 3

Using sum-of-logs = log-of-product:

$$ g(x) = \sum_{n=1}^p \ln{(1+nx)} = \ln(\prod_{n=1}^p (1+nx)) $$ $$ h(x) = e^{g(x)} = \prod_{n=1}^p (1+nx) $$ and $(\ln a)^b = b\ln a$: $$ f(x) = g(x)^x = (\ln h(x))^x = x \ln h(x) $$

As $x\rightarrow 0$, each term in $h(x)$ approaches $1$, so $h(x)\rightarrow 1$, and $$ \lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0} x \ln h(x) = 0 \times \ln(1) = 0\times 0 = 0. $$